11
$\begingroup$

Given a path integral for a system

$$Z(\phi) = \int [D\phi] e^{-S[\phi]},$$

where I am working in the Euclidean signature, necessarily mean that the system described is quantum mechanical? In the equation above, I am looking at $(d+1)$ dimensional field theories, such that $d=0$ Feynman path integral is standard quantum mechanics, and $d>0$ Feynman path integral means QFTs. Periodicity in the Euclidean time similarly gives the thermal partition function of the quantum system.

To pose this question more precisely, are there path integrals of the above form corresponding to which there exist no canonical formulation of quantum mechanics. A provocative example might be the Martin-Siggia-Rose stochastic path integral, which seemingly admits no quantum description. However it is dual to a microscopic description of a Brownian particle interacting with a bath if viewed in the Schwinger Keldysh formalism. Thus it defines a quantum system. My question is do all path integrals admit a description in the canonical quantum mechanical formulation, be it a direct or an indirect dual interpretation; or are there obstructions to such interpretations for certain classes of path integrals?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ What is your definition of "quantum mechanical" to have a common ground of discussion. Some would say you need $\hbar$ in your path integral and that lowest orders in an expansion in powers of $\hbar$ correspond to classicality... Alternatively we could say a system is quantum mechanical if typical wavelengths involved are of the same size as the compton wavelengths of the particles involved... $\endgroup$ – ohneVal Jul 8 at 15:02
  • $\begingroup$ @ohneVal The standard formulation of quantum mechanics, where we have physical observables described by operators acting on Hilbert space. $\endgroup$ – Bruce Lee Jul 8 at 15:05
  • $\begingroup$ I am not speaking about different forumlations of quantum mechanics, I am asking about classicality vs quantum effects. As I understand from the post your concern is about quantum mechanical effects. $\endgroup$ – ohneVal Jul 8 at 15:07
  • $\begingroup$ @ohneVal My question is given a path integral, does it always admit a description via the canonical formulation. Maybe this was not clear, thanks for pointing this out. $\endgroup$ – Bruce Lee Jul 8 at 15:10
  • $\begingroup$ Suggested title (v2): Does a path integral formulation imply an operator formalism? $\endgroup$ – Qmechanic♦ Jul 8 at 15:16
2
$\begingroup$

One set of sufficient conditions for the path integral to generate a Wightmanian Quantum Field Theory in the Minkowski space is known as Osterwalder-Schrader axioms.

Of course the real problem is to actually give precise mathematical meaning to the formal path integral, and then to establish that the resulting correlation functions satisfy these axioms.

In practice, the axiom that is usually the most non-trivial and difficult to check is reflection positivity. It translates into unitarity after the Wick rotation to Wightmanian QFT.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Very interesting, but isn't this about the Wick rotation specifically? The path integral formalism is not the same as the Wick rotation, although the latter is often used within it. The OP should make a clearer statement on what he means exactly $\endgroup$ – ohneVal Jul 8 at 15:27
  • $\begingroup$ @ohneVal Well, the OS reconstruction theorem can be thought of as a mathematically precise formulation of the general intuition behind the Wick rotation. $\endgroup$ – Prof. Legolasov Jul 8 at 15:28
  • $\begingroup$ Yes I agree completely, I am just saying that I don't believe Wick rotation is the main point of the OP's question, but perhaps I am wrong $\endgroup$ – ohneVal Jul 8 at 15:29
  • $\begingroup$ The point you make is correct. However, it doesn't address my question. $\endgroup$ – Bruce Lee Jul 8 at 15:29
  • $\begingroup$ @BruceLee perhaps I misunderstood your question then. It seems to me you are asking about conditions under which a path integral leads to quantum mechanics / field theory. If that is indeed what you are asking, the OS reconstruction theorem guarantees that if the OS axioms are satisfied by the correlation functions (which are path integrals), then there is a corresponding QFT (or QM in $d=1$) in Minkowski space with the Hilbert space, field operators, the Hamiltonian, etc. $\endgroup$ – Prof. Legolasov Jul 8 at 15:31
1
$\begingroup$

Let us first try to agree on definitions. The path integral has essentially two ingredients, an action and a set of boundary conditions.

A given action specifies the model and can used under very different frameworks to obtain trajectories, observables or other relevant quantities of a model.

For some actions, where one is interested in quantum mechanical behavior, one must additionally quantize the system, canonically or other methods. The Feynman path integral is (meaning the path integral and its boundary conditions) another method for doing that. Namely as you know, both ways lead to the same correlation functions which is what we measure, they both describe the same physics which is quantum mechanical.

The wording "quantum mechanical" should be taken as suggested by @Qmechanic to be operator formalism. For which you appear to already give an example.

At the end of the day it is up to the action and what it is trying to describe, not about the path integral. If you have an action which allows canonical quantization, the path integral will yield the same results. If you are given an arbitrary action for an arbitrary system the answer is it cannot always be first quantized consistently (without suffering some problem e.g. "unboundedness" from below).

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

First of all, path integrals are used beyond quantum theory and even beyond physics - I am thinking, first of all, about Onsager-Matchlup functional used for diffusive systems, and widely applied in Finance.

Path integrals usually arise as an alternative to a probabilistic description in terms of a partial differential equation or Langevin equations. I cannot make an exact statement that there is a PDE corresponding to any path integral, but this well may be true within the limits relevant to physical theories.

In physics the alternative is often between using Feynmann-Dyson expansion and path integral formulation, which are equivalent, but differ by how easily certain types of approximations are made.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I understand your point. However my question is, is there any obstruction to such path integrals defining a quantum mechanical system? $\endgroup$ – Bruce Lee Jul 8 at 15:07
  • $\begingroup$ The question is then about what actually makes a system quantum mechanical: one is the nature of the small parameter, $\hbar$? The presence of imaginary unit (which reflects the relationship between space and time)? The type of the underlying differential equation? $\endgroup$ – Vadim Jul 8 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.